Question

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation \[ 2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}, \] is ______ .

Answer 1

Correct Answer: 0

We are given the equation:

\[2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}\]

Let \(\sin^{-1} x = \alpha\) and \(\cos^{-1} x = \beta\). We know the identity:

\[\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\]

So, we have:

\[2\alpha + 3\beta = \frac{2\pi}{5}\]

Using \(\beta = \frac{\pi}{2} - \alpha\), substitute this into the equation:

\[2\alpha + 3\left(\frac{\pi}{2} - \alpha\right) = \frac{2\pi}{5}\]

Simplifying:

\[2\alpha + \frac{3\pi}{2} - 3\alpha = \frac{2\pi}{5}\]

\[-\alpha + \frac{3\pi}{2} = \frac{2\pi}{5}\]

\[-\alpha = \frac{2\pi}{5} - \frac{3\pi}{2}\]

\[-\alpha = \frac{4\pi}{10} - \frac{15\pi}{10} = -\frac{11\pi}{10}\]

\[\alpha = \frac{11\pi}{10}\]

Now, since \(\alpha = \sin^{-1} x\) and \(\sin^{-1} x\) must lie in the range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we find that \(\alpha = \frac{11\pi}{10}\) is not possible because it is outside the allowed range of the inverse sine function.

Thus, the equation has no real solutions.

Final Answer:

0